# Is Wick rotation necessary?

1. Jan 26, 2012

### atyy

Is Wick rotation something that is necessary to define QFT, or is it just a calculational trick that is a quick way to get a result that is justified by some other means?

2. Jan 26, 2012

### Ben Niehoff

A Wick rotation is a quick-and-dirty way to circumvent some mathematical subtleties, as far as I can tell, but I can't say I've ever seen these subtleties made explicit.

3. Jan 26, 2012

### ParticleGrl

Wick rotation's are used in a variety of different ways, and can mean slightly different things. Are you talking about the replacement of it/hbar with 1/kT? Or simply using t -> itau in order to move a Minkowski metric to a Euclidean one? The latter is a substitution to make the math a bit easier.

4. Jan 26, 2012

### atyy

I was thinking of something like what's described on p112 of Srednicki's http://web.physics.ucsb.edu/~mark/ms-qft-DRAFT.pdf . It looks like we'd like to know an integral along the infinite real line. To evaluate it we treat as as part of a contour integral. Then he says that the integral stays the same if we rotate it, since it doesn't pass through poles.

Is the original integral well-defined?

Although rotating the contour doesn't change its value, how do we know that integral along the infinite imaginary line has the same value as the original integral along the infinite real line?

5. Jan 26, 2012

### ParticleGrl

We know BECAUSE rotating the contour doesn't change the value.

Its a standard trick from complex analysis. Start by promoting the integral along the real line to a contour integral (we can do this if the function limits to 0 fast enough as you head outward on the complex plane). We can then use the standard residue theorem to evaluate the integral.

In this case, instead of using residue theorem, we just make use of the fact that the result of the contour integral just depends on the pole structure, so as long as we don't cross any poles when we rotate the contour, we know we don't change the value, so the two integrals are equivalent.

6. Jan 27, 2012

### atyy

I guess the curve that closes the contour has the same value in both cases?

Also, in the discussion on p113 he says the original integral (14.18) is divergent unless d<4, but after Wick rotation it's finite for d<8. If the original integral isn't divergent, I guess Wick rotation is just a calculational trick. But in the cases where the original integral isn't well-defined, is Wick rotation more than a calculational trick and actually a defining assumption?

7. Jan 27, 2012

### ParticleGrl

Yes, to turn an infinite line integral into a contour integral without altering the value, the value of the function at large distance had best be small enough that taking the limit that your closing contour moves out to infinity had better give you 0.

Wick rotation won't change the dimensions where the integral diverges. If you look carefully, you'll see that he has taken derivatives of the original integral.

If you have

$$\int \frac{1}{k^{2m}} d^D k$$

Your integral will go like $k^{D-2m}$. If D is larger than 2m, this will blow up on you. Now, what happens if you take derivatives with respect to k^2?

Also, yes, the procedure where you take derivatives to render a divergent integral finite, do the integral, and then integrate to "undo" the original derivatives is something that makes field theory hard to deal with rigorously. There is a formally infinite integration constant kicking around somewhere. This isn't the wick rotation though, its a combination of a regulator and some renormalization.

Last edited: Jan 27, 2012
8. Jan 27, 2012

### Ben Niehoff

The Wick rotation in the context given does depend on the pole prescription, but the pole prescription comes from the definition of the Feynman propagator. The poles on the real axis get moved up and down in just such a way that the Wick rotation makes sense.

In my earlier post, I was thinking more of the idea of going to Euclideanized time to make the path integral appear more convergent. This is where the mathematical subtleties come up. For example, in Euclideanized time, one can find Yang-Mills instantons by solving self-dual equations. But the existence of self-dual 2-forms relies on the Euclidean signature of R^4; in Minkowski space, self-dual forms do not exist. So there is some subtlety in exactly how these instantons enter the path integral.

9. Jan 27, 2012

### atyy

I see. So in this context once we put the iε the Wick rotation is justified.

So is the Wick rotation also justified in the path integral case? I'm a little familiar with Wilsonian renormalization in classical statistical mechanics where one sums over configurations in the canonical ensemble. My understanding of the classical stat mech case is that if the theory has a cut-off, and a symmetry below the cut-off, then renormalization generates all terms consistent with the symmetry as one flows to lower energies, where there is a fixed point if the experimentalist puts the sample near the critical temperature. I've have a vague understanding that the same ideas apply in QFT, and have imagined that one just makes time imaginary and it's the same. Is this imaginary time also a Wick rotation, and is it similarly well-defined already in Minkowski spacetime before making the replacement?

10. Jan 27, 2012

### tom.stoer

The Wick rotation can be defined rather rigorously once the pole and cut structure is known. This is possible in QM, but b/c exact solutions are out of reach in QFT, it is by no means clear whether Wick rotation is a valid operation; in addition it cannot be justified based on the results which may be valid in a restricted domain only (e.g. perturbatively); one may miss certain features of the theory.

Anyway - b/c there are no mathematically rigorous constructions of quantum field theories, Wick rotation is just another dirty trick.

11. Jan 29, 2012

### atyy

I still don't understand how the Wick rotation is justified in renormalization, eg. Hollowood's notes: "We take it as established fact that one can move between the Minkowski and Euclidean versions of the theory without difficulty."

Googling seems to indicate I should try learning about the Matsubara formalism. Is this right, or am I barking up the wrong tree?

12. Jan 29, 2012

### atyy

I looked up some books by Collins, Rivasseau and Salmhofer. They all seem to agree that the Wick rotation is required as an assumption in the definition of QFT via a path integral in Minkowski spacetime. Salmhofer says otherwise the kinetic term is not semi-bounded.

In that case, the Wick rotation is primary and the iε prescription derived. The derivation of imaginary time via the Matsubara formalism seems applicable only for equilibrium states.

Salmhofer further says that Wilsonian renormalization starts with a theory that has a cut-off in momentum space, not via a lattice discretization (I suppose unless the theory is asymptotically free or safe and does not require a cut-off), otherwise one cannot analytically continue from Euclidean to Minkowski spacetime.

Rivasseau mentions that the Euclidean theory must satisfy the Osterwalder-Schrader conditions or some alternatives in order to anaytically continue from Euclidean to Minkowski spacetime. He also mentions it's hard to show that all such conditions are met.

The only dissenting view seems to be from Cartier and DeWitte-Morette who still try to define the path integral directly in Minkowski spacetime without Wick rotation, and consider the rotation derived rather than primary.

Last edited: Jan 29, 2012
13. Jan 30, 2012

### tom.stoer

so they turn it bround and use it as a 'definition'. OK, I mean the whole PI stuff is not well-defined afaik, so it's no problem to add another ingredient ...

14. Jan 30, 2012

### atyy

Hmm, maybe it's alright. Googling around keeps turning up the opinion that the analytic continuation exists if the Osterwalder-Schrader conditions are fulfilled.

http://www.einstein-online.info/spotlights/path_integrals (QM)
http://arxiv.org/abs/hep-lat/9807028 (Lattice QCD)

BTW, I've only seen Kadanoff-Wilson renormalization (cut-off with a lattice and block spins, or equivalent heuristic in momentum space) with path integrals. Can Kadanoff-Wilson renormalization be done in the canonical formalism?

15. Jan 30, 2012

### tom.stoer

I think that's the way they studied it originally in condensed matter physics

16. Jan 31, 2012

### atyy

Thanks! I looked up the Kondo effect papers and that's indeed how they did it.

17. Feb 2, 2012

### Physics Monkey

Regarding the validity of the rotation, it may be true in some formal sense that things are analytic in the appropriate way and can be continued. Nevertheless, actually doing the continuation is an enormous practical problem. For example, at finite temperature in a quantum many-body system it may be most convenient to formulate the problem in Euclidean signature. The physical response functions of interest are then obtained by analytic continuation of the Euclidean correlators, however, it is known that this continuation does not commute with many common approximations including the epsilon and large N expansions. Sometimes this is physically obvious since, for example, large N is like a thermodynamic limit and hence can introduce new singularities. In these situations one must often formulate the epsilon or large N expansion directly in real time at considerable complication e.g. in a Keldysh or quantum Boltzmann kind of setup.

The problem of analytic continuation also affects Euclidean lattice studies of field theory. It may be very hard to extract dynamical quantities at finite temperature such as the viscosity.

18. Feb 2, 2012

### atyy

Is it right to say that at zero temperature, all quantities, including dynamical quantities, can be calculated at least in principle by analytically continuation from imaginary time, but that at finite temperature, some dynamical quantities cannot even be in principle calculated from imaginary time?